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# The locus of the poles of normal chords of the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ isa) $\dfrac{{{x^2}}}{{{a^4}}} + \dfrac{{{y^2}}}{{{b^4}}} = {a^2} + {b^2}$  b) $\dfrac{{{x^2}}}{{{a^4}}} + \dfrac{{{y^2}}}{{{b^4}}} = {a^2} - {b^2}$ c) $\dfrac{{{a^6}}}{{{x^2}}} + \dfrac{{{b^6}}}{{{y^2}}} = {({a^2} - {b^2})^2}$ d) $\dfrac{{{a^4}}}{{{x^2}}} + \dfrac{{{b^4}}}{{{y^2}}} = {({a^2} - {b^2})^2}$

Last updated date: 18th Jun 2024
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Hint: The given equation is the equation of an ellipse. We will take a polar coordinate, and put it in the given equation by substituting only one $x$ and $y$ by the taken polar coordinate. Then we will find out the normal to the ellipse and compare the generated equation with the standard form of the normal of an ellipse, by certain algebraic manipulations we will get the equation for locus of the poles of normal chords of the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ .
Formula used:
1) $\csc \theta = \dfrac{1}{{\sin \theta }},\sec \theta = \dfrac{1}{{\cos \theta }}$
2) ${\sin ^2}\theta + {\cos ^2}\theta = 1$

We are given in the question that the equation of the ellipse is
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
So the given equation is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ --------(i)
Now we will try to take a polar coordinate to substitute in one $x$ and $y$ in the above equation (i)
Let $\left( {h,k} \right)$ be the polar coordinate.
Now polar equation of $\left( {h,k} \right)$ with respect to the ellipse is given by
$\dfrac{{xh}}{{{a^2}}} + \dfrac{{yk}}{{{b^2}}} = 1$ --------(ii)
If the above equation represents normal to the ellipse then it must be identical to the general form of the equation of normal of an ellipse.
$ax\sec \theta - by\csc \theta = {a^2} - {b^2}$
Comparing this with (ii) we get:
$\dfrac{{\left( {\dfrac{h}{{{a^2}}}} \right)}}{{a\sec \theta }} = \dfrac{{\left( {\dfrac{k}{{{b^2}}}} \right)}}{{ - b\csc \theta }} = \dfrac{1}{{{a^2} - {b^2}}}$
$\dfrac{{{a^2} - {b^2}}}{a}\left( {\dfrac{h}{{{a^2}}}} \right) = \sec \theta$ and $\dfrac{{{a^2} - {b^2}}}{{ - b}}\left( {\dfrac{k}{{{b^2}}}} \right) = \csc \theta$
$\Rightarrow \cos \theta = \dfrac{a}{{{a^2} - {b^2}}}\left( {\dfrac{{{a^2}}}{h}} \right)$ and $\sin \theta = \dfrac{{ - b}}{{{a^2} - {b^2}}}\left( {\dfrac{{{b^2}}}{k}} \right)$
$\Rightarrow \cos \theta = \dfrac{{\left( {\dfrac{{{a^3}}}{h}} \right)}}{{{a^2} - {b^2}}}$ and $\sin \theta = \dfrac{{ - \left( {\dfrac{{{b^3}}}{k}} \right)}}{{{a^2} - {b^2}}}$
Now, squaring both terms and adding them we get:
${\cos ^2}\theta + {\sin ^2}\theta = {\left[ {\dfrac{{\left( {\dfrac{{{a^3}}}{h}} \right)}}{{{a^2} - {b^2}}}} \right]^2} + {\left[ {\dfrac{{ - \left( {\dfrac{{{b^3}}}{k}} \right)}}{{{a^2} - {b^2}}}} \right]^2}$
$\Rightarrow 1 = \dfrac{1}{{{{({a^2} - {b^2})}^2}}}\left[ {\left( {\dfrac{{{a^6}}}{{{h^2}}}} \right) + \left( {\dfrac{{{b^6}}}{{{k^2}}}} \right)} \right]$
On cross multiplication we get:
$\left( {\dfrac{{{a^6}}}{{{h^2}}}} \right) + \left( {\dfrac{{{b^6}}}{{{k^2}}}} \right) = {({a^2} - {b^2})^2}$ Which is the required locus for $\left( {h,k} \right)$
Replacing $\left( {h,k} \right)$ with $\left( {x,y} \right)$ we get: $\left( {\dfrac{{{a^6}}}{{{x^2}}}} \right) + \left( {\dfrac{{{b^6}}}{{{y^2}}}} \right) = {({a^2} - {b^2})^2}$
So, the correct answer is “Option C”.

Note: The algebraic substitutions and calculation involves so many complicated terms, use only one operation at a time. Applying operations repeatedly in the same process may result in errors in your solution.